I noted a discussion on groups being abelian under a certain restriction on powers of elements, e.g. http://tiny.cc/chs45. Maybe this result (probably not too well-known) concludes it all.

Let and $m$ and $n$ be coprime natural numbers. Assume that $G$ is a group such that $m$-th powers commute and $n$-th powers commute (that is for all $g, h$ $\in$ $G$: $g^mh^m=h^mg^m$ and $g^nh^n=h^ng^n$). Then $G$ is abelian.

@Derek: Let $G$ be a finite group. Assume that there are two coprime integers $m$ and $n$ such that for all $g, h \in G$ holds (1) $g^mh^m = h^mg^m$ and (2) $g^nh^n = h^ng^n$. How do you prove that $G$ is abelian?
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SomeoneSep 27 '11 at 14:06

G does not have to be finite. Let $M \subset G$ be the subgroup generated by all m-th powers and let $N \subset G$ be the subgroup generated by all n-th powers. These subgroups are clearly abelian normal subgroups. Since m and n are coprime $G = MN$ and $M \cap N$ is contained in the center $Z(G)$ of $G$.
To prove that G is abelian it suffices to show that M and N commute, that is $[M,N]=1$. Note that $[M,N] \subset (M \cap N)$. Let $a \in M$ and $b \in N$.
Then $[a, b] = a^{−1}b^{−1}ab \in M \cap N$. Hence $[a, b] = z$ with $z \in Z(G)$. Hence $b^{−1}ab = za$, whence $b^{−1}a^nb=z^na^n$. Since $a^n \in N$ it commutes with $b$, so $z^n=1$. Similarly $z^m=1$. Since $m$ and $n$ are relatively prime, we conclude $z=1$.